(a) What is the tangential acceleration of a bug on the rim of a 12.0-in.-diamet
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Question
(a) What is the tangential acceleration of a bug on the rim of a 12.0-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 76.0 rev/min in 4.50 s? (b) When the disk is at its final speed, what is the tangential velocity of the bug? (c) One second after the bug starts from rest, what is its tangential acceleration? (d) One second after the bug starts from rest, what is its centripetal acceleration? (e) One second after the bug starts from rest, what is its total acceleration?Explanation / Answer
The tangential speed is vt = ?*r (a) vt0 = ?0*r vt1 = ?1*r ?vt = vt1 - vt0 = (?1 - ?0)*r tangential acceleration at = ?vt/?t ?0 = 0, ?1 = 76 RPM = 7.96 rad/s r = 6 in = 0.152 m ?vt = 7.96*0.152 = 1.21 m/s at = 1.21/3 = 0.40 m/s² (b) vt1 = ?1*r = 1.21 m/s (c) if the angular acceleration is uniform, then the tangential acceleration is constant. the centripetal acceleration is ?²*r. after 1 s, the angular velocity is 1/3 the final velocity at 3 s. so ? = 7.96/3 = 2.65 rad/s and the centripetal acceleration is 2.65²*0.152 = 1.07 m/s² The total acceleration is the sq rt of the sum of the squares of centripetal and tangential accelerations: a = v[1.07² + 0.40²] = 1.14 m/s² The angle rel to at is arctan (ac/at) = arctan (1.07/0.4) = 69.5º
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